Solve (2+3i)div(1+2i)

$$( 2 + 3 i ) \div ( 1 + 2 i )$$

Evaluate

$$\frac{8}{5}-\frac{1}{5}i=1.6-0.2i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-2i.

$$\frac{\left(2+3i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2+3i\right)\left(1-2i\right)}{1^{2}-2^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{\left(2+3i\right)\left(1-2i\right)}{5}$$

Multiply complex numbers 2+3i and 1-2i like you multiply binomials.

$$\frac{2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)i^{2}}{5}$$

By definition, i^{2} is -1.

$$\frac{2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)\left(-1\right)}{5}$$

Do the multiplications in 2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)\left(-1\right).

$$\frac{2-4i+3i+6}{5}$$

Combine the real and imaginary parts in 2-4i+3i+6.

$$\frac{2+6+\left(-4+3\right)i}{5}$$

Do the additions in 2+6+\left(-4+3\right)i.

$$\frac{8-i}{5}$$

Divide 8-i by 5 to get \frac{8}{5}-\frac{1}{5}i.

$$\frac{8}{5}-\frac{1}{5}i$$

Real Part

$$\frac{8}{5} = 1\frac{3}{5} = 1.6$$

Solution Steps

Multiply both numerator and denominator of \frac{2+3i}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$Re(\frac{\left(2+3i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{\left(2+3i\right)\left(1-2i\right)}{1^{2}-2^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{\left(2+3i\right)\left(1-2i\right)}{5})$$

Multiply complex numbers 2+3i and 1-2i like you multiply binomials.

$$Re(\frac{2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)i^{2}}{5})$$

By definition, i^{2} is -1.

$$Re(\frac{2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)\left(-1\right)}{5})$$

Do the multiplications in 2\times 1+2\times \left(-2i\right)+3i\times 1+3\left(-2\right)\left(-1\right).

$$Re(\frac{2-4i+3i+6}{5})$$

Combine the real and imaginary parts in 2-4i+3i+6.

$$Re(\frac{2+6+\left(-4+3\right)i}{5})$$

Do the additions in 2+6+\left(-4+3\right)i.

$$Re(\frac{8-i}{5})$$

Divide 8-i by 5 to get \frac{8}{5}-\frac{1}{5}i.

$$Re(\frac{8}{5}-\frac{1}{5}i)$$

The real part of \frac{8}{5}-\frac{1}{5}i is \frac{8}{5}.

$$\frac{8}{5}$$